Problem: Simplify and expand the following expression: $ \dfrac{n}{4n - 3}+\dfrac{-6}{3n + 3} $
Answer: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(4n - 3)(3n + 3)$ Multiply the first term by $\dfrac{3n + 3}{3n + 3}$ $ \begin{align*} \dfrac{n}{4n - 3} \times \dfrac{3n + 3}{3n + 3} & = \dfrac{(n)(3n + 3)}{(4n - 3)(3n + 3)} \\ & = \dfrac{3n^2 + 3n}{(4n - 3)(3n + 3)}\end{align*} $ Multiply the second term by $\dfrac{4n - 3}{4n - 3}$ $ \begin{align*} \dfrac{-6}{3n + 3} \times \dfrac{4n - 3}{4n - 3} & = \dfrac{(-6)(4n - 3)}{(3n + 3)(4n - 3)} \\ & = \dfrac{-24n + 18}{(3n + 3)(4n - 3)}\end{align*} $ Now we have: $ = \dfrac{3n^2 + 3n}{(4n - 3)(3n + 3)} + \dfrac{-24n + 18}{(3n + 3)(4n - 3)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{3n^2 + 3n - 24n + 18}{(4n - 3)(3n + 3)} $ $ = \dfrac{3n^2 - 21n + 18}{(4n - 3)(3n + 3)}$ Expand the denominator: $ = \dfrac{3n^2 - 21n + 18}{12n^2 + 3n - 9}$ Simplify: $ = \dfrac{n^2 - 7n + 6}{4n^2 + n - 3}$